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Absolutely continuous spectrum, Krein systems, Sturm-Liouville  
operators.
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\begin{document}

\title{On the existence of the absolutely continuous component for the measure
associated with some orthogonal systems. }
\date{{\small {\it Moscow State University, Faculty of Computational Mathematics
and Cybernetics, Vorob'evy Gory, Moscow, Russia, 119899. e-mail address:
saden@cs.msu.su\\}}}
\author{Sergey A. Denisov}
\maketitle

\begin{abstract}
In this article, we consider two orthogonal systems: Sturm-Liouville
operators and Krein systems. For Krein systems, we study the behavior of
generalized polynomials at the infinity for spectral parameters in the upper
half-plain. That makes it possible to establish the presence of absolutely
continuous component of the associated measure. For Sturm-Liouville operator
on the half-line with bounded potential $q$, we prove that essential support
of absolutely continuous component of the spectral measure is $[m,\infty)$
if $\limsup_{x\to \infty} q(x)=m$ and $q^\prime\in L^2(R^+)$. That holds for
all boundary conditions at zero. This result partially solves one open
problem stated recently by S.Molchanov, M.Novitskii, and B.Vainberg.
\end{abstract}

{\ }

\begin{center}
{\bf \S 0. Introduction. }
\end{center}

The contents of the paper is as follows. In the first paragraph, we prove
the asymptotics of generalized polynomials for Krein systems with
coefficients of a special kind. We also establish the presence of absolutely
continuous component of the associated measure. In the second paragraph,
results obtained for Krein system are applied to Sturm-Liouville operators.

In this introductory paragraph, we will remind some results for the Krein
systems. The Krein systems are defined by the equations

\begin{equation}
\left\{ 
\begin{array}{cccc}
\frac{dP(x,\lambda )}{dx} & = & i\lambda P(x,\lambda )-\overline{A(x)}%
P_{*}(x,\lambda ), & \,P(0,\lambda)=1, \\ 
\frac{dP_{*}(x,\lambda )}{dx} & = & -A(x)P(x,\lambda ), & \,
P_*(0,\lambda)=1,
\end{array}
\right.  \label{sys3}
\end{equation}
where $A(x)$ is locally summable function on $R^+$.

In his famous work \cite{Kr}, M.G.Krein showed that $P(x,\lambda)$ have many
properties of polynomials orthogonal on the unit circle \footnote{%
The spectral theory of Krein systems was developed further in \cite{Akh,
Den, Ryb, Sakh1, Sakh2, Sakh3}.}. For example, there exists a non-decreasing
function $\sigma(\lambda)$ (spectral measure ) defined on the whole line
such that mapping $U_P(f)=\int\limits_0^\infty f(x)P(x,\lambda)dx$ is
isometry from $L^2(R^+)$ to $L^2(\sigma,R)$. We will call $P(x,\lambda)$ the
generalized polynomials. The following Theorem was stated in \cite{Kr}. It
was proved later in \cite{Sakh1}.

{\bf Theorem \cite{Kr,Sakh1}.} {\it \ The following statements are equivalent%
}

{\it (1) The integral $\int\limits_{-\infty }^\infty \frac{\ln \sigma
^{^{\prime }}(\lambda )}{1+\lambda ^2}d\lambda$ is finite.}

{\it (2) At least at some }$\lambda $, $\Im \lambda >0$, {\it the integral } 
\begin{equation}
\int\limits_0^\infty \left| P(x,\lambda )\right| ^2dx  \label{int3}
\end{equation}

{\it converges.}

{\it (3) At least at some }$\lambda, ${\it \ }$\Im \lambda >0$,{\it \ the
function }$P_{*}(x,\lambda )${\it \ is bounded.}

{\it (4) On any compact set in the open upper half-plane, integral {\rm (\ref
{int3})} converges uniformly. That is equivalent to the existence of uniform
limit }$\Pi (\lambda )=\lim_{x\rightarrow \infty }P_{*}(x,\lambda ).${\it \ }






Consider some measure $\mu$ on $R$. Let $I$ be the finite union of intervals
on $R$. Let us assume that for any measurable set $\Omega\in I$ with
positive Lebesgue measure ($|\Omega|>0$ ), we have $\mu (\Omega )>0$. That
already means that $\mu$ has nontrivial absolutely continuous component.
Though not any measure with nontrivial absolutely continuous component has
this property. If this condition holds, we say that the essential support of
absolutely continuous component of measure $\mu$ is $I$ ($essupp\{ \mu_{ac}
\}=I$).

Thus, if one of the conditions {\it (2)-(4)} holds, then the essential
support of $\sigma_{ac}(\lambda)$ is $R$. It is easy to show \cite{Kr} that $%
A(x)\in L^2(R)$ or $A(x)\in L^1(R)$ yields {\it (3)}. In \cite{Den}, we
proved the criterion for {\it (3)} to hold in terms of coefficients $A(x)$
from the so-called Stummel class. In the next paragraph, we will study
another class of perturbations that does not shrink the essential
support for absolutely continuous component
of the measure. The similar problems for Sturm-Liouville operators were studied in
numerous publications (see, for example, \cite{MNV} and bibliography there
). But first, let us outline some relations between Krein systems and some
other orthogonal systems. Consider the Dirac system

\begin{equation}
\left\{ 
\begin{array}{cccc}
\phi^\prime & = & -\lambda\psi-a_1\phi+a_2\psi, & \phi(0)=1, \\ 
\psi^\prime & = & \lambda\phi+a_2\phi+a_1\psi, & \psi(0)=0,
\end{array}
\right.  \label{dir}
\end{equation}
where $a_1=2\Re A(2x)$, $a_2=2\Im A(2x)$. It turns out that $e^{-i\lambda
x}P(2x,\lambda)=\phi(x,\lambda)+i\psi(x,\lambda)$. That allows us to say 
\cite{Kr} that $\rho_{Dir}(\lambda)=2\sigma(\lambda)$, where $%
\rho_{Dir}(\lambda)$-- spectral measure of Dirac systems (\ref{dir}).

In case $a_2=0$ and $a_1$-- absolutely continuous, we have

\begin{equation}
\begin{array}{ccc}
\psi^{\prime\prime}-q\psi+\lambda^2\psi=0, & \psi(0)=0, & 
\psi^\prime(0)=\lambda, \\ 
\phi^{\prime\prime}-q_1\phi+\lambda^2\phi=0, & \phi(0)=1, & 
\phi^\prime(0)+a_1(0)\phi(0)=0,
\end{array}
\end{equation}
where $q=a_1^2+a_1^\prime,\ q_1=a_1^2-a_1^\prime$.

Therefore, the spectral measure $\rho_d(\lambda)$ of Sturm-Liouville
operator $l(u)=-u^{\prime\prime}+qu$ with Dirichlet boundary condition $%
u(0)=0$ is related to $\sigma(\lambda)$ by 
\begin{equation}
\rho_d (\lambda)=4\int\limits_0^{\sqrt\lambda} \xi^2d\sigma(\xi),\ \lambda>0.
\label{zero}
\end{equation}

{\ }

\begin{center}
{\bf \S 1. Krein systems with bounded coefficient which derivative is square
summable. }
\end{center}

In this paragraph, we will prove the following Theorem.

{\bf Theorem 1.} {\it If bounded coefficient $A(x)$ of Krein system is real
valued,} $0<l_1=\liminf_{x\rightarrow \infty }A(x)\leq \limsup_{x\rightarrow
\infty }A(x)=l_2$,{\it \ and } $A^{\prime }\in L^2(R^{+})$, {\it then } $%
(-\infty ,-2l_2]\cup [2l_2,\infty )\subset essupp\{\sigma _{ac}\}$.

But first we will prove one auxiliary Lemma. Consider the Krein system

\begin{equation}
\left\{ 
\begin{array}{cccc}
P^\prime & = & i\lambda P-A P_*, & P(0)=1, \\ 
P_*^\prime & = & -AP, & P_*(0)=1,
\end{array}
\right.
\end{equation}

where real valued $A$ is bounded, $0< l_1=\liminf_{x\rightarrow \infty} A
\leq \limsup_{x\rightarrow \infty }A=l_2$, and $A^{\prime }\in L^2(R^{+})$.

Matrix $\aleph=\left[ 
\begin{array}{cc}
i\lambda & -A \\ 
-A & 0
\end{array}
\right]$ has eigenvalues $\displaystyle \mu_1=\frac{i\lambda-\sqrt{%
4A^2-\lambda^2}}{2}$, $\displaystyle \mu_2=\frac{i\lambda+\sqrt{%
4A^2-\lambda^2}}{2}$. Consider $\lambda=\tau+ik$, $\tau>2l_2$ is fixed, $k>0 
$. Assume for simplicity $A$ is such that $l_1<2A<\tau/2+l_2$ for all $x\in
R^+$. We will explain later why this assumption can always be made without
loss of generality. Symbol $C$ is reserved for positive constants which
value might change from one formula to another.  It is easy to
verify that $\Re \mu_1>0$ for all $x>0, k>0$.

The following Lemma is true.

{\bf Lemma.} {\it The asymptotics holds at the infinity }

{\it 
\begin{equation}
P_*(x,k)=\exp\left[ \int\limits_0^x \mu_1(s,k) ds \right] O(x,k),
\end{equation}
}

{\it where $|O(x,k)|<\exp(C/k)$ for all $x>0, k>0$}. 

{\it Proof.}
Many estimates in this proof are very crude, but they will be good 
enough for our purposes. Let $J$ be $2\times2$ matrix that satisfies equation 
$J^\prime=\aleph J$. We
will find $J$ in the form $J=LQ$, where $L=\left[ 
\begin{array}{cc}
-\mu_1/A & -\mu_2/A \\ 
1 & 1
\end{array}
\right] $ consists of eigenvectors of $\aleph$. We have the following
equation for $Q$: $Q^\prime=L^{-1}\aleph LQ-L^{-1}L^\prime Q$. Multiplying
matrixes, we have

\begin{equation}
Q^\prime=\left[ 
\begin{array}{cc}
\mu_1 & 0 \\ 
0 & \mu_2
\end{array}
\right]Q+VQ,
\end{equation}
where

\begin{equation}
V=\left[ 
\begin{array}{cc}
v_{11} & v_{12} \\ 
v_{21} & v_{22}
\end{array}
\right] = \frac{1}{\sqrt{4A^2-\lambda^2}} \left[ 
\begin{array}{cc}
-\frac{A^\prime}{A}\mu_1+\mu_1^\prime & -\frac{A^\prime}{A}\mu_2+\mu_2^\prime
\\ 
\frac{A^\prime}{A}\mu_1-\mu_1^\prime & \frac{A^\prime}{A}\mu_2-\mu_2^\prime
\end{array}
\right].
\end{equation}

Let us notice that the following inequality 
\begin{equation}
\displaystyle \int\limits_0^\infty \|V\|^2 dx<C  \label{star}
\end{equation}
holds uniformly in $k$. Introduce matrix

\begin{equation}
Q_\circ= \left[ 
\begin{array}{cc}
\exp\Bigl[ \int\limits_0^x (\mu_1(s,k)+v_{11}(s,k)) ds\Bigr] & 0 \\ 
0 & \exp\Bigl[ \int\limits_0^x (\mu_2(s,k)+v_{22}(s,k)) ds \Bigr]
\end{array}
\right].
\end{equation}

If $Q=Q_\circ S$, then for $S=\left[ 
\begin{array}{cc}
s_{11} & s_{12} \\ 
s_{21} & s_{22}
\end{array}
\right] $, $S(0)=I$, we have

\begin{equation}
S^\prime =\left[ 
\begin{array}{cc}
0 & v_{12}\exp \Bigl[ \int\limits_0^x \nu(s,k) ds \Bigr] \\ 
v_{21} \exp \Bigl[ -\int\limits_0^x \nu(s,k) ds \Bigr] & 0
\end{array}
\right]S,
\end{equation}
where $\displaystyle \nu=\sqrt{4A^2 -\lambda^2}+ \frac{i\lambda A^\prime}{A%
\sqrt{4A^2-\lambda^2}}$.

For $s_{11}$ and $s_{21}$, we have

\begin{equation}
\left\{ 
\begin{array}{cccc}
s_{11}^\prime & = & v_{12}\exp \Bigl[ \int\limits_0^x \nu(s,k) ds \Bigr]
s_{21}, & s_{11}(0)=1, \\ 
s_{21}^\prime & = & v_{21}\exp \Bigl[ -\int\limits_0^x \nu(s,k) ds \Bigr]
s_{11}, & s_{11}(0)=0.
\end{array}
\right.  \label{sys1}
\end{equation}

Therefore, for $s_{11}$, we have integral inequality

\begin{equation}
|s_{11}|\leq 1+ \int\limits_0^x |s_{11}(t)||v_{21}(t)| \int\limits_t^\infty
|v_{12}(s)| |\exp \Bigl[ \int\limits_t^s \nu(\xi) d\xi \Bigr] | ds dt.
\label{ie}
\end{equation}

Notice that $\displaystyle 
\Bigl| \int\limits_t^s \frac{i\lambda A^{\prime}}{A\sqrt{4A^2-\lambda^2}}%
d\xi \Bigr|<C$ uniformly in $k,t,s$.

Because $\Re \mu_1>0$, we have the estimate

\begin{equation}
\Re \sqrt{4A^2-\lambda^2}<-k.
\end{equation}

Consequently, Gronwall Lemma, being applied to (\ref{ie}), yields

\begin{equation}
|s_{11}|\leq \exp\left\{ C \int\limits_0^x |v_{21}(t)| \int\limits_t^\infty
|v_{12}(s)| \exp \Bigl( -{k[s-t]} \Bigr) ds dt \right\} \leq \exp(C/k).
\end{equation}
At the final step, we used (\ref{star}) and the Young inequality for
convolutions.

Therefore, for $s_{21}$, we have the estimate

\[
|s_{21}|\leq \exp(C/k) \int\limits_0^x |v_{21}(s)| |\exp \Bigl[
-\int\limits_0^s \nu(\xi,k) d\xi \Bigr]|ds
\]
which follows from the second equation of system (\ref{sys1}) and estimate
on $s_{11}$. In the same way, the estimates for $s_{12}, s_{22}$ can be
obtained. They are as follows

\begin{equation}
|s_{12}|\leq \frac{C}{\sqrt k} \exp \left( \frac{C}{k} \right),
\end{equation}

\begin{equation}
|s_{22}|\leq \frac{C}{\sqrt k} \exp \left( \frac{C}{k} \right)
\int\limits_0^x |v_{21}(s)| |\exp \Bigl[ -\int\limits_0^s \nu(\xi,k) d\xi
\Bigr]|ds.
\end{equation}

If $J$ is such that $J(0)=I$, then $J=LQ_\circ S L^{-1}(0)$. Therefore, for $%
P_*$, we will have

\begin{equation}
P_*=\exp \Bigl[ \int\limits_0^x (\mu_1(t)+v_{11}(t)) dt\Bigr] \Bigl\{ \alpha
s_{11}+ \beta s_{12}+\Bigr.
\end{equation}

\begin{equation}
\Bigl. \exp\left( \int\limits_0^x (\mu_2(s)-\mu_1(s)+v_{22}(s)-v_{11}(s))ds
\right) (s_{21}\alpha+s_{22}\beta) \Bigr\}.
\end{equation}

Constants $\alpha$ and $\beta$ are chosen in such a way that the initial
condition $P_*(0,k)=1$ is satisfied. We have

\begin{equation}
\left[ 
\begin{array}{c}
\alpha \\ 
\beta
\end{array}
\right] =\frac 1{\sqrt{4A^2(0)-\lambda ^2}}\left( 
\begin{array}{cc}
A(0) & \mu _2(0) \\ 
-A(0) & -\mu _1(0)
\end{array}
\right) \times \left[ 
\begin{array}{c}
1 \\ 
1
\end{array}
\right] ,
\end{equation}

therefore, $\alpha$ and $\beta$ are bounded uniformly in $k$. Denote by $%
O(x,k)$

\begin{equation}
O=\exp \Bigl[ \int\limits_0^x v_{11}(t) dt\Bigr] \Bigl\{ \alpha s_{11}+
\beta s_{12}+\Bigr.  \label{ooo}
\end{equation}

\begin{equation}
\Bigl. \exp\left( \int\limits_0^x (\mu_2(s)-\mu_1(s)+v_{22}(s)-v_{11}(s))ds
\right) (s_{21}\alpha+s_{22}\beta) \Bigr\}.
\end{equation}

Notice that $\displaystyle \int\limits_0^x v_{11}(t) dt$ and $\displaystyle %
\int\limits_0^x v_{22}(t) dt$ are bounded uniformly in $x$ and $k$. Due to
the estimates on $s_{21}$ and $s_{22}$, we have 
\begin{equation}
\Bigl| \exp\left( \int\limits_0^x (\mu_2(s)-\mu_1(s)+v_{22}(s)-v_{11}(s))ds
\right) (s_{21}\alpha+s_{22}\beta) \Bigr|\leq  \label{ff1}
\end{equation}

\begin{equation}
\exp(C/k) \left| \exp \Bigl[ \int\limits_0^x \nu(\eta,k) d\eta \Bigr]
\right| \times
\end{equation}

\begin{equation}
\int\limits_0^x |v_{21}(s)| \left| \exp \Bigl[- \int\limits_0^s \nu(\xi,k)
d\xi \Bigr] \right| ds \leq
\end{equation}

\begin{equation}
\exp(C/k) \int\limits_0^x |v_{21}(s)| \left| \exp \Bigl[ \int\limits_s^x
\nu(\xi,k) d\xi \Bigr] \right| ds \leq \exp(C/k)/{\sqrt k}.  \label{ff2}
\end{equation}

To get the last estimate, we used Cauchy inequality. Finally, bounds on $%
s_{11}$, $s_{12}$ lead to $|O(x,k)|<\exp (C/k)$. That finishes the proof of
the Lemma. $\Box$

{\it Remark. } More accurate estimates on $\alpha$, $\beta$, $s_{11}$ allow
us to write inequality

\begin{equation}
|O(x,z)-1|<1/2,  \label{ret}
\end{equation}

where $x\in R^+, \tau \leq \Re z \leq \tau +1, \Im z>k_0$, $k_0$-- some
positive constant.

Indeed, it is easy to verify that $\alpha\to 1$ and $\displaystyle |\beta|<%
\frac{C}{\Im z}$ if $\Im z\to +\infty$, $\tau\leq \Re z \leq \tau+1$. From (%
\ref{sys1}), by Cauchy inequality we have

\begin{equation}
|s_{11}-1|\leq \exp(C/k)/k
\end{equation}
uniformly in $x\in R^+$, $\tau \leq \Re z \leq \tau+1$, where $k=\Im z>k_0$.

One can verify that $\displaystyle 
\int\limits_0^x v_{11}(t) dt\to 0$ uniformly in $x$ if $\Im z\to \infty$.

Therefore, from (\ref{ooo}) and (\ref{ff1})-(\ref{ff2}), we infer (\ref{ret}%
).

Let us prove Theorem 1 now.

{\it Proof of Theorem 1.}
Fix any $\tau >2l_2$. We will show that $[\tau ,\infty )\subset essupp\{\sigma
_{ac}\}$. Because $\tau $ is chosen arbitrary larger than $2l_2$ and $\sigma 
$ is odd (since $A$ is real ), this inclusion is sufficient for 
Theorem 1 to be true. Once $%
\tau $ is fixed, we can assume that $l_1<2A<\tau/2 +l_2$ for all $x\in R^{+}$%
. Due to the standard trace-class perturbation argument applied to Dirac
system (\ref{dir}) \cite{RS3}, we can always make this assumption. Indeed, 
since $%
l_1=\liminf_{x\rightarrow \infty}A\leq \limsup_{x\rightarrow \infty }A=l_2$
as $x\to \infty $, it suffices to multiply $A$ by some smooth function
which is equal to $0$ on $[0,M_A]$ and $1$ on $[M_A+1,\infty )$. The
absolutely continuous part of $\rho _{Dir}$ will not change because of
trace-class argument. On the other hand, if $M_A$ is sufficiently large, we
will satisfy the imposed condition.

Consider Krein system with coefficients $\displaystyle
A_{(n)}(x)=\left\{ 
\begin{array}{cc}
A(x), & x<n, \\ 
0, & x>0.
\end{array}
\right. $

Denote the corresponding measure by $\sigma_n$. We have the formula ((3.14)
from [4] ):

\begin{equation}
\sqrt{2\pi} P_{*}(\infty,z)=\sqrt{2\pi} P_{*}(n,z)= e^{i\alpha_n} \exp
\left[ \int\limits_{-\infty}^{\infty} \frac{1}{2\pi i} \frac{(1+tz) \ln
\sigma_n^\prime (t)} {(z-t)(1+t^2)}dt \right], \Im z>0,  \label{rew}
\end{equation}
where $\alpha_n$ are some real constants. Because $A(x)$ is real, functions $%
\sigma_n$ are odd. Therefore, if we take $z=i$, the left-hand side of (\ref
{rew}), together with exponent from the right-hand side, are real valued.
Thus, $\alpha_n$ can be chosen equal to zero.

The asymptotics of $P_{*}(n,z)$ as $|z|\to \infty$ is $P_{*}(n,\infty)=1$ 
\cite{Ryb}. Therefore, we can rewrite this formula as follows

\begin{equation}
-2\pi i \ln P_*(n,z) = \int\limits_{-\infty}^{\infty} \frac{(1+tz) \ln (2\pi
\sigma_n^\prime (t))} {(t-z)(1+t^2)}dt  \label{ttt}
\end{equation}

if $\Im z >0$. Here we used the identity

\[
\sqrt{2\pi }=\exp \left[ \int\limits_{-\infty}^{\infty} -\frac{1}{2\pi i} 
\frac{(1+tz) \ln(2\pi) } {(z-t)(1+t^2)}dt \right],\ \Im z>0. 
\]

Also we used the fact that $2\pi \sigma_n^\prime(t)\to 1$ if $|t|\to \infty$ 
\cite{Ryb}.

The right-hand side $RHS(z)$ of (\ref{ttt}) satisfies the condition $%
\overline{RHS(z)}=RHS(\bar{z})$. Let us define the left-hand side for $\Im
z<0$ according to this rule. So we have some analytic function $LHS(z)$
defined in the region: $\Im z\neq 0$. Recall the asymptotic formula for $%
P_*(x,z)$, $(\Im z>0, \Re z=\tau>2l_2 )$:

\begin{equation}
P_*(x,z)=\exp\left[ \int\limits_0^x \mu_1(s,z) ds \right] O(x,z).
\end{equation}

Then,

\[
LHS(z)=-2\pi i \int\limits_0^n \mu_1(s,z) ds -2\pi i\ln O(n,z) 
\]

for $\Im z>0$.

It is easy to see that the continuation according to the chosen rule for
function $-2\pi i \int\limits_0^n \mu_1(s,z) ds $ is exactly the Schwarz
analytic continuation. It follows from the fact that the value of this
function on the half-line $\Im z=0, \Re z\geq \tau$ is real.

Consider $z=\tau+ik$. Let $\ln O(n,z)=r_1(n,k)+ir_2(n,k)$. From the
definition of $LHS(z)$ in $\Im z<0$, we have $\overline{i(r_1(n,k)+ir_2(n,k)}%
=-ir_1(n,k)-r_2(n,k)=ir_1(n,-k)-r_2(n,-k)$. Thus, $r_1(n,k)$ is odd and $%
r_2(n,k)$ is even. For fixed $n$, they are well defined functions for all $k
\neq 0$ with finite right and left limits at $k=0$. Notice also that for $k>0
$, $r_1(n,k)=\ln |O(n,\tau+ik)|$ and $r_2(n,k)=Arg O(n,\tau+ik)$. For $r_1$, 
$r_2$,\ we have $r_1(n,\infty)=0$,\ $r_2(n,\infty)=0$ because $%
P_*(n,\infty)=1$ and $\mu_1 (s,\infty)=0$.

Then, integrate the both sides of (\ref{ttt}), together with some auxiliary
function $\displaystyle  \frac{(z-\tau)^3}{((z-\tau)^2+m^2)^4}$, along the
contour $Con$. We choose $Con$ as the contour that consists of the complex
numbers $\tau+ik$ ($|k|\leq m-1, |k| \geq m+1$ ) and two right semicircles
with radii $1$ centered at $\tau \mp im$. The direction of integration is
upward.

For the right-hand side, we have

\[
\int\limits_{Con} \frac{(z-\tau)^3}{((z-\tau)^2+m^2)^4} \int\limits_{-%
\infty}^{\infty} \frac {(1+tz)\ln (2\pi \sigma_n^\prime (t))}
{(1+t^2)(t-z)}dt dz= 
\]

\[
\int\limits_{-\infty}^{\infty} \frac{\ln (2\pi \sigma_n^\prime (t))} {1+t^2}
\int\limits_{Con} \frac{(1+tz)(z-\tau)^3} {((z-\tau)^2+m^2)^4 (t-z) } dz dt=
2\pi i \int\limits_{\tau}^\infty \frac{(t-\tau)^3} {((t-\tau)^2+m^2)^4} \ln
(2\pi \sigma_n^\prime (t))dt. 
\]

Here we changed the order of integration by Fubini Theorem and then used the
Cauchy formula.

Integrating the $LHS(z)$ with the same function, we have

\[
-2\pi i \int\limits_{Con} \int\limits_0^n \mu_1(s,z) ds \frac{(z-\tau)^3}{%
((z-\tau)^2+m^2)^4}dz -2\pi i \int\limits_{Con} \frac{(z-\tau)^3}{%
((z-\tau)^2+m^2)^4} \ln O(n,z)dz= 
\]
\[
0-2\pi i \int\limits_{Con} \frac{(z-\tau)^3}{((z-\tau)^2+m^2)^4} \ln
O(n,z)dz 
\]
because $\int\limits_0^n \mu_1(s,z) ds$ is analytic and bounded 
in $\Re z \geq \tau$.

Thus, we have the equality

\begin{equation}
- \int\limits_{Con} \frac{(z-\tau)^3}{((z-\tau)^2+m^2)^4} \ln O(n,z) dz=
\int\limits_{\tau}^\infty \frac{(t-\tau)^3}{((t-\tau)^2+m^2)^4} \ln (2\pi
\sigma_n^\prime (t))dt.  \label{ger}
\end{equation}

We have the following inequality \cite{Sakh1}

\begin{equation}
\int\limits_{-\infty}^\infty \frac{d\sigma_n(t)}{1+t^2}<C  \label{pp}
\end{equation}

uniformly in $n$. Define $\ln^+ h =\ln h$ for $h>1$ and zero otherwise, $%
\ln^- h= -\ln h$ if $0<h<1$ and zero otherwise. The trivial inequality $%
\ln^+ h <h $, together with (\ref{pp}), guarantee that the right-hand side
of (\ref{ger}) is bounded above uniformly in $n$.

Our goal is to show that it is bounded from below as well.

Thus, we want to show that

\[
\int\limits_{Con} \frac{(z-\tau)^3}{((z-\tau)^2+m^2)^4} \ln O(n,z) dz < C 
\]
uniformly in $n$.

The left hand side is equal to $I_1$ +$I_2$. Where

\[
I_1=\int\limits_{|k|<m-1, |k|>m+1} i\frac {(ik)^3}{(m^2-k^2)^4}
(r_1(n,k)+ir_2(n,k))dk= 
\]

\[
2\int\limits_{0<k<m-1, k>m+1} \frac{k^3}{(m^2-k^2)^4} r_1(n,k)dk \leq C 
\]

uniformly in $n$ due to the estimates on $\ln |O(n,k)|$. We also used the
fact that $r_2(n,k)$ is even. It is crucial since we do not have control
over the argument of $O(n,k)$. We only know the upper bound on $\ln |O(n,k)|$%
. Term $I_2$ corresponds to the integration along the semicircles. Choose $m$
sufficiently large. Then, for $I_2$, we have the estimate

\[
|I_2|\leq C 
\]

uniformly in $n$ because of the estimate (\ref{ret}).

To finish the proof, we will use one argument from \cite{DK}. Consider any
compact set $Co\in (\tau, \infty)$ of positive Lebesgue measure. Then, as it
follows from (\ref{pp}) and boundedness of right-hand side in (\ref{ger}), $%
\int\limits_{Co} \ln^- \sigma_n^\prime (t) dt$ is bounded uniformly in $n$.
Jensen's inequality yields

\[
\ln^- \left\{ \frac{1}{|Co|}\int\limits_{Co} \sigma_n^\prime(t)dt \right\}
\leq \frac{1}{|Co|} \int\limits_{Co} \ln^- \sigma_n^\prime (t)dt. 
\]

Therefore, $\sigma_n(Co)$ is greater than some positive constant $d(Co)$ for
all $n$. The Weyl-Titchmarsh function of system with coefficients $A_{(n)}$
converges to Weyl-Titchmarsh function of system with coefficient $A$ \cite
{Sakh1}. This convergence is uniform on any compact set of upper half-plain.
By Stone-Weierstrass Theorem, we have weak convergence of $\sigma_n$ to $%
\sigma$. Consequently, $\sigma(Co)\geq \limsup _{n\to \infty} \sigma_n(Co)
>0 $ for each compact $Co$ with $|Co|>0$. $\Box$

{\ }

\begin{center}
{\bf \S 2. Sturm-Liouville operators. }
\end{center}

The main goal of this paragraph is to prove the following Theorem. The idea
of the proof will follow \cite{Den1} more or less. But it will require more
technical details.

{\bf Theorem 2. } {\it Consider Sturm-Liouville operator on the half-line
given by differential expression }

{\it 
\begin{equation}
l(u)=-u^{\prime\prime}+qu,  \label{sl}
\end{equation}
}

{\it where $q$ is bounded, }$\limsup_{x\rightarrow \infty }q=m,${\it \ $q(x)$
is absolutely continuous, and $q^{\prime }\in L^2(R^{+})$. Then, for any
boundary condition at zero, the essential support of absolutely continuous
component of the corresponding spectral measure is }$[m,\infty )${\it .}

The problems of this kind for Sturm-Liouville operators were treated in
many publications. We mention some of the results obtained. P.Deift and
R.Killip proved

{\bf Theorem \cite{DK}.} {\it If in (\ref{sl}), $q(x)\in L^2(R^+)$, then the
essential support of absolutely continuous component of spectral measure is $%
R^+$ for all boundary conditions.}

In the paper of S.Molchanov, M.Novitskii, and B.Vainberg \cite{MNV}, some
other results were obtained. For example, it was proved that $q\in
L^3(R^{+}) $ and $q^{\prime }\in L^2(R^{+})$ lead to the same property of
the spectral measure. Consider the particular case of Theorem 2, where $q\to 0$
as $x\to\infty$, $q^{\prime}\in L^2(R^+)$. We see that condition $q\in
L^3(R^+)$ used in \cite{MNV} can be relaxed to $q\to 0$ as $x\to \infty$. In 
\cite{MNV}, authors also pose the following open problem. Is it true that
for all boundary conditions at zero, $essupp \{ \rho_{ac} \} =[m,\infty )\ $
provided that $q$ is bounded, $\ \limsup_{x\rightarrow \infty }q=m$, and for
some integer $p$, $q^{(p)}\in L^2(R^{+})$. Here $q^{(p)}$ means the
derivative of order $p$.

Theorem 2 solves this problem for $p=1$. Because any bounded function admits
supremum, we actually characterize the absolutely continuous component of
the spectrum for bounded potentials with square summable first derivative.
We think that method developed here can be used to deal with any $p$. No
doubts, it will require more calculations to establish the asymptotics.

Let us prove the following Lemma first.

{\bf Lemma.} {\it If bounded }$q${\it \ is such that }$\limsup_{x\rightarrow
\infty }q=m${\it \ and $q^{\prime }\in L^2(R^{+})$, then for some large $%
\gamma >0$, there exists bounded $v(x)$ so that }$\limsup_{x\rightarrow
\infty }v=\sqrt{\gamma ^2+m}-\gamma $, {\it \ $v^{\prime }\in L^2(R^{+})$,
and $q=v^2+2\gamma v+v^{\prime }$.}

{\it Proof }. Consider the corresponding integral equation

\begin{equation}
v(x)=e^{-2\gamma x} \int\limits_0^x e^{2\gamma s} (q(s)-v^2(s))ds.  \label{m1}
\end{equation}

If $v(x)$ is solution of this integral equation, then it satisfies
differential equation as well. Write (\ref{m1}) as follows $v=OP_\gamma v$,
where $OP_\gamma $ is corresponding formal nonlinear operator. Consider the
complete metric space $M=\{f\in C(R^{+}),\Vert f\Vert _\infty \leq 1\}$ with 
$\Vert .\Vert _\infty $ metric. By $C(R^{+})$ we denote continuous functions
on $R^{+}$. If $\gamma $ is large enough, operator $OP_\gamma $ acts from $M$
to $M$. Naturally, the choice of $\gamma$ depends on $\|q\|_\infty$. For
large $\gamma $, $OP_\gamma $ has contracting property. Indeed,

\[
|OP_\gamma g_1 -OP_\gamma g_2 | \leq e^{-2\gamma x} \int\limits_0^x
e^{2\gamma s} |g_1-g_2| |g_1+g_2|ds\leq \frac 12 \|g_1-g_2\|_\infty, 
\]

for $\gamma$ large enough. Therefore, there is the unique fixed point from $%
M $. Let us call this function $v$.

Differentiate ($\ref{m1}$). After integration by parts, we will have

\[
v^\prime =e^{-2\gamma x}(q(0)-v^2(0))+e^{-2\gamma x}\int\limits_0^x
e^{2\gamma s} q^\prime (s)ds-2e^{-2\gamma x}\int\limits_0^x e^{2\gamma s}
v(s)v^\prime (s)ds. 
\]

Consider integral equation

\begin{equation}
b =e^{-2\gamma x}(q(0)-v^2(0))+e^{-2\gamma x}\int\limits_0^x e^{2\gamma s}
q^\prime (s)ds-2e^{-2\gamma x}\int\limits_0^x e^{2\gamma s} v(s)b(s)ds.
\label{m2}
\end{equation}

It has the unique solution from $C(R^+)$. Therefore, its solution is $%
v^\prime$. The uniqueness follows from the convergence of the corresponding
iterated series. Write (\ref{m2}) as $b=OP1_\gamma b$.

\[
e^{-2\gamma x}\int\limits_0^x e^{2\gamma s} q^\prime (s)ds=\int_0^\infty
r_\gamma (x-s) q^\prime(s)ds, 
\]

where $r_\gamma(s)=e^{-2\gamma s}$ for positive $s$, and zero otherwise.
Because $q^\prime\in L^2(R^+)$, the Young inequality for convolutions yields
that $OP1_\gamma$ acts from ball $\|.\|_2\leq 1$ into itself, provided that $%
\gamma$ is large enough. For large $\gamma$, it has contracting property.
Therefore, there is the unique fixed point $b$ in ball $\|.\|_2\leq 1$
which is equal to $v^\prime$. We also used the fact that $\|v\|_\infty \leq
1 $. It is clear that $v^\prime (x)\to 0$ as $x \to \infty$. Therefore,
solving equation $v^2+2\gamma v+v^\prime-q=0$ gives formula

\begin{equation}
v=-\gamma \pm \sqrt{\gamma^2 -v^\prime +q}.  \label{ame}
\end{equation}

We know that $q$ is bounded and $\|v\|_\infty\leq 1$ for large $\gamma>0$.
Consequently, to obtain asymptotics of $v$ at the infinity, one should take
sign $+$ in (\ref{ame}). Therefore, we have $\limsup_{x\to \infty} v(x)
=-\gamma+\sqrt{\gamma^2+m}$. $\Box$

Now the proof of Theorem 2 is straightforward.

{\it Proof of the Theorem 2.} Notice that the essential support of the
absolutely continuous component does not depend on the boundary condition at
zero. It follows, for example, from the subordinacy theory \cite{GP,Sim}.
Consider (\ref{sl}) with Dirichlet boundary condition at zero. Add to the
potential $q$ some constant $\gamma ^2$. Denote the corresponding
self-adjoint operator by $H_\gamma $. Obviously, the spectral measure
 of $H_\gamma $
is the shift of the spectral measure of the initial operator $H_0$. On the other hand,
for large $\gamma $, we can solve equation $q+\gamma ^2=(v+\gamma
)^2+(v+\gamma )^{\prime }$. That is due to Lemma of this paragraph. Now we
can apply results of the first paragraph for Krein systems with coefficient $%
A(x)=\gamma /2+v(x/2)/2$. We can use Theorem 1 because for large $\gamma$, $%
\displaystyle 0<\liminf_{x\to \infty} A(x) \leq \limsup_{x\to \infty} A(x) = 
\frac{\sqrt{\gamma^2+m}}{2}$. Let us use formula (\ref{zero}) from the
Introduction. Thus, we see that $[\gamma ^2+m,\infty )\subset essupp \{\rho
_{ac}(H_\gamma )\}$ which is equivalent to $[m,\infty )\subset essupp
\{\rho _{ac}(H_0)\}$.

To prove that it is actually an equality, one should use some result from 
\cite{St} which goes back to \cite{SS}. Before formulating this result, let
us introduce some notations used in \cite{St}. Consider $V$ -- real valued
and locally integrable on $(0,\infty ).$ Assume that $-\frac{d^2}{dx^2}+V$
is limit point at $+\infty $ and $\int\limits_0^\varepsilon |V(s)|ds<\infty
. $ Let $T=-\frac{d^2}{dx^2}+V$ in $L^2(R^{+})$ with any fixed boundary
condition at zero. Consider also $V_{\circ }$ -- integrable and bounded from
below function. Denote by $T_{\circ }$ operator generated by $-\frac{d^2}{%
dx^2}+V_{\circ }$ in $L^2(R).$ Then, the following theorem holds.

{\bf Theorem (\cite{St}, Theorem 2.1 ).} {\it Assume that }$-\infty \leq
\alpha <\beta \leq +\infty \ ${\it and }$(\alpha ,\beta )\cap Spectrum
(T_{\circ })=\emptyset \ ${\it and that intervals }$I_n\subset [0,\infty )$%
{\it \ exist such that }

\[
|I_n|\rightarrow \infty \ as\ n\rightarrow \infty \ and\ \sup_{x\in \cup
_nI_n}|V(x)-V_{\circ }(x)|\leq \delta . 
\]

{\it Then, } $essupp \{ \rho _{ac}(T) \} \cap (\alpha +\delta ,\beta -\delta
)=\emptyset .$

Now, consider $V_{\circ }=m$, $V=q$. Because $\limsup_{x\to \infty }q=m$ and 
$q^{\prime }\in L^2(R^{+})$, one can easily show that for any $\delta >0$, $%
essupp \{ \rho _{ac}(H_0) \} \cap (-\infty ,m-\delta )=\emptyset $. Indeed,
it suffices to choose $x_n\to \infty $ such that $|q(x_n)-m|<\delta/2 $ and $%
I_n$ as some neighboring intervals. We specify them as follows. Notice that $%
q(x)-q(x_n)=\int\limits_{x_n}^x q^{\prime }(s)ds$. Therefore,

$|q(x)-q(x_n)|\leq \left( \int\limits_{x_n}^\infty \left[ q^{\prime
}(s)\right] ^2ds\right) ^{1/2}\cdot \sqrt{x-x_n}=\varepsilon _n\sqrt{x-x_n}, 
$

where $\varepsilon _n\rightarrow 0$ as $n\rightarrow \infty $. Choose as $%
I_n $ intervals $[x_n,x_n+\varepsilon _n^{-1}]$ for $n$ so large that $%
\varepsilon_n<\frac{\delta^2}{4}$. Evidently, $|I_n|\rightarrow \infty $ and
for each $x\in I_n$, we have $|q(x_n)-m|<\delta $. Therefore, Stolz Theorem
yields $essupp \{\rho _{ac}(H_0)\}=[m,\infty ).\Box $

{\it Remark.} There are many functions that satisfy conditions of Theorem 2.
In particular, these are some slowly oscillating functions. One can think
about $\cos (x^\mu )$ for $0<\mu <1/2$. In that case $\limsup_{x\to \infty
}q(x)=1$, $q^{\prime }\in L^2(R^{+})$. Consequently, $essupp \{ \rho _{ac}
\} =[1,\infty )$ for all boundary conditions. In paper \cite{St},
author used the theory of subordinate solutions to show that for any $0<\mu
<1$, the spectrum is actually purely absolutely continuous on $[1,\infty )$.
The interval $[-1,1]$ is filled by singular spectrum.

{\it Remark.} The condition $q^{\prime }\in L^2(R^{+})$ from Theorem 2 is
optimal \cite{MNV}. That means that the statement can be false if $%
q^{\prime}\in L^p(R^{+}),\ p>2$. Famous von Neumann-Wigner potential \cite
{vNW,RS4} satisfies the conditions of the Theorem 2. That means that under the
conditions of Theorem 2, the singular component of spectrum can appear on the
interval $[m,\infty )$ which supports the absolutely continuous part.

{\it Remark.} It should be noted that class of Sturm-Liouville
operators with decaying potentials match very well the class of Krein
systems with coefficients that tend to nonzero constant. For example,
Sturm-Liouville operators of this kind admit negative eigenvalues with zero as the only
possible point of accumulation. For the Krein systems, these eigenvalues
of the discrete spectrum 
might accumulate only near the edges of the corresponding symmetric
interval centered at zero. 
This relation can be explained as follows. Consider potential from
$L^p(R^+), (1\leq p<\infty )$ space and Dirichlet boundary condition 
for example. Then, for $\gamma$ large enough, equation
(\ref{m1}) can be solved so that $v\in L^\infty (R^+)\cap L^p(R^+)$.
Consequently, formula (\ref{zero}) from the Introduction allows us to study
Krein system with coefficient $A=\gamma/2+v(x/2)/2$ instead of initial 
Sturm-Liouville operator.


Methods, developed in this paper, can be applied to some other classes of
Sturm-Liouville operators and Krein systems. For example, using the same
arguments, one can prove that the result of P.Deift and R.Killip holds for
more general class of potentials. Namely, for uniformly square summable
functions from $H^{-1}(R^+)$ space, i.e.

\[
\int\limits_x^{x+1} q^2(s)ds < C 
\]

uniformly in $x\in R^+$, and

\[
e^{-x} \int\limits_0^x e^s q(s)ds \in L^2(R^+). 
\]

That allows to treat some oscillations at the infinity. This result solves
one open problem stated in \cite{Den1}. The proof is essentially the same as
the proof of Theorem 2. Instead of studying the Sturm-Liouville operator,
we consider the corresponding Krein system with coefficient 
$A(x)=\gamma/2+v(x/2)/2$, where bounded $v(x)$ is from $L^2(R^+)$. The methods
from Theorem 1 are applied then. 

{\it Acknowledgment.} Author thanks professor J.Goodman for useful
discussion. Most of this work was done during the stay at Courant Institute of
Mathematical Sciences. The work was partially supported by RFBR
 Grant N00-15-96104.

\newcounter{count} \setcounter{count}{2}

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